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that there exists no trace on ?(H) if H is an //zmte-dimensional Hilbert space
(Problem at the end of this section).

The corresponding notion for positive Hilbert forms is that of a bitrace.
A bitrace on A is by definition a positive Hilbert form g on A which satisfies
the additional condition

(15.7.3)                                  S(y*,x) = g(x9y).

This reduces to (1 5.7.1 ) when g(x, y) =f(y*x), /being a positive linear form.


(15J.4) Consider the algebra ^2(E) of Hilbert-Schmidt operators on an
infinite-dimensional separable Hilbert space (15.4.8). For any Hilbert
basis (#) of E and any pair (u, v) of Hilbert-Schmidt operators on E, the sum
 (u(0n) | v(anj) is defined, because (u(an) \ v(an)) = ((v*u)(an) \ an\ and we have

seen that the double family ((v*u)(an) \ am) is absolutely summable (15.4.8),
since v*u is a Hilbert-Schmidt operator. Moreover, for each Hilbert
basis (bn) of E, we have

) = I (() I bm)(bn | t,(an))

and therefore

(b J |

This shows that the sum J] (u(an) \ v(an)) does not depend on the orthonormal


basis (an) originally chosen. If we denote this sum by g(u, v), then it is
immediately seen that g is a positive hermitian form on J5?2(E), such that
g(u,u)= \\u\\2- Also, if w, v, w are three Hilbert-Schmidt operators, then
(u(v(an)) | w(aw)) = (r(aw) | u*(w(ajj)9 hence ^(wu, w) = g(v, M*W); and the cal-
culation above shows that g(v*, w*) = #(w, y). Hence g is a bitrace on ^f 2(E).
It can be shown that it does not come from a trace on this algebra (Problem).
Note also that g satisfies conditions (U) and (N) of (15.6). For (U), this fol-
lows from the inequality (; for (N), observe that the endomorphisms
of finite rank are dense in JS?2(E) (15.4.8) and that if w(E) = F is finite-
dimensional then we may write u = PF o u (in the notation of (6.3.1)) where
PF is also of finite rank.other hand, it is easily seen a (not necessarily